NoteNextra-origin/content/Math401/Extending_thesis/Math401_S4.md

Math 401, Fall 2025: Thesis notes, S4, Complex function spaces and complex manifold

Bargmann space (original)

Also known as Segal-Bargmann space or Bargmann-Fock space.

It is the space of holomorphic functions that is square-integrable over the complex plane.

Section belows use Remarks on a Hilbert Space of Analytic Functions as the reference.

A family of Hilbert spaces, \mathfrak{F}_n(n=1,2,3,\cdots), is defined as follows:

The element of \mathfrak{F}_n are entire analytic functions in complex Euclidean space \mathbb{C}^n. f:\mathbb{C}^n\to \mathbb{C}\in \mathfrak{F}_n

Let f,g\in \mathfrak{F}_n. The inner product is defined by

\langle f,g\rangle=\int_{\mathbb{C}^n} \overline{f(z)}g(z) d\mu_n(z)

Let z_k=x_k+iy_k be the complex coordinates of z\in \mathbb{C}^n.

The measure \mu_n is the defined by

d\mu_n(z)=\pi^{-n}\exp(-\sum_{i=1}^n |z_i|^2)\prod_{k=1}^n dx_k dy_k

Example

For n=2,

\mathfrak{F}_2=\text{ space of entire analytic functions on } \mathbb{C}^2\to \mathbb{C}

\langle f,g\rangle=\int_{\mathbb{C}^2} \overline{f(z)}g(z) d\mu(z),z=(z_1,z_2)

d\mu_2(z)=\frac{1}{\pi^2}\exp(-|z|^2)dx_1 dy_1 dx_2 dy_2

so that f belongs to \mathfrak{F}_n if and only if \langle f,f\rangle<\infty.

This is absolutely terrible early texts, we will try to formulate it in a more modern way.

The section belows are from the lecture notes Holomorphic method in analysis and mathematical physics

Complex function spaces

Holomorphic spaces

Let U be a non-empty open set in \mathbb{C}^d. Let \mathcal{H}(U) be the space of holomorphic (or analytic) functions on U.

Let f\in \mathcal{H}(U), note that by definition of holomorphic on several complex variables, f is continuous and holomorphic in each variable with the other variables fixed.

Let \alpha be a continuous, strictly positive function on U.

\mathcal{H}L^2(U,\alpha)=\left\{F\in \mathcal{H}(U): \int_U |F(z)|^2 \alpha(z) d\mu(z)<\infty\right\},

where \mu is the Lebesgue measure on \mathbb{C}^d=\mathbb{R}^{2d}.

Theorem of holomorphic spaces

1. For all z\in U, there exists a constant c_z such that

   
   |F(z)|^2\le c_z \|F\|^2_{L^2(U,\alpha)}
   

   for all F\in \mathcal{H}L^2(U,\alpha).
2. \mathcal{H}L^2(U,\alpha) is a closed subspace of L^2(U,\alpha), and therefore a Hilbert space.

Proof

First we check part 1.

Let z=(z_1,z_2,\cdots,z_d)\in U, z_k\in \mathbb{C}. Let P_s(z) be the "polydisk"of radius s centered at z defined as

P_s(z)=\{v\in \mathbb{C}^d: |v_k-z_k|<s, k=1,2,\cdots,d\}

If z\in U, we cha choose s small enough such that \overline{P_s(z)}\subset U so that we can claim that F(z)=(\pi s^2)^{-d}\int_{P_s(z)}F(v)d\mu(v) is well-defined.

If d=1. Then by Taylor series at v=z, since F is analytic in U we have

F(v)=F(z)+\sum_{k=1}^{\infty}a_n(v-z)^n

Since the series converges uniformly to F on the compact set \overline{P_s(z)}, we can interchange the integral and the sum.

Using polar coordinates with origin at z, (v-z)^n=r^n e^{in\theta} where r=|v-z|, \theta=\arg(v-z).

For n\geq 1, the integral over P_s(z) (open disk) is zero (by Cauchy's theorem).

So,

\begin{aligned}
F(z)&=(\pi s^2)^{-1}\int_{P_s(z)}F(z)+\sum_{k=1}^{\infty}a_n(v-z)^n d\mu(v)\\
&=(\pi s^2)^{-1}F(z)+(\pi s^2)^{-1}\sum_{k=1}^{\infty}a_n\int_{P_s(z)}r^n e^{in\theta} d\mu(v)\\
&=(\pi s^2)^{-1}F(z)
\end{aligned}

For d>1, we can use the same argument to show that

Let \mathbb{I}_{P_s(z)}(v)=\begin{cases}1 & v\in P_s(z) \\0 & v\notin P_s(z)\end{cases} be the indicator function of P_s(z).

\begin{aligned}
F(z)&=(\pi s^2)^{-d}\int_{U}\mathbb{I}_{P_s(z)}(v)\frac{1}{\alpha(v)}F(v)\alpha(v) d\mu(v)\\
&=(\pi s^2)^{-d}\langle \mathbb{I}_{P_s(z)}\frac{1}{\alpha},F\rangle_{L^2(U,\alpha)}
\end{aligned}

By definition of inner product.

So \|F(z)\|^2\leq (\pi s^2)^{-2d}\|\mathbb{I}_{P_s(z)}\frac{1}{\alpha}\|^2_{L^2(U,\alpha)} \|F\|^2_{L^2(U,\alpha)}.

All the terms are bounded and finite.

For part 2, we need to show that \forall z\in U, we can find a neighborhood V of z and a constant d_z such that

|F(z)|^2\leq d_z \|F\|^2_{L^2(U,\alpha)}

Suppose we have a sequence F_n\in \mathcal{H}L^2(U,\alpha) such that F_n\to F, F\in L^2(U,\alpha).

Then F_n is a cauchy sequence in L^2(U,\alpha). So,

\sup_{v\in V}|F_n(v)-F_m(v)|\leq \sqrt{d_z}\|F_n-F_m\|_{L^2(U,\alpha)}\to 0\text{ as }n,m\to \infty

So the sequence F_m converges locally uniformly to some limit function which must be F (\mathbb{C}^d is Hausdorff, unique limit point).

Locally uniform limit of holomorphic functions is holomorphic. (Use Morera's Theorem to show that the limit is still holomorphic in each variable.) So the limit function F is actually in \mathcal{H}L^2(U,\alpha), which shows that \mathcal{H}L^2(U,\alpha) is closed.

which shows that \mathcal{H}L^2(U,\alpha) is closed.

Tip

[1.] states that point-wise evaluation of F on U is continuous. That is, for each z\in U, the map \varphi: \mathcal{H}L^2(U,\alpha)\to \mathbb{C} that takes F\in \mathcal{H}L^2(U,\alpha) to F(z) is a continuous linear functional on \mathcal{H}L^2(U,\alpha). This is false for ordinary non-holomorphic functions, e.g. L^2 spaces.

Reproducing kernel

Let \mathcal{H}L^2(U,\alpha) be a holomorphic space. The reproducing kernel of \mathcal{H}L^2(U,\alpha) is a function K:U\times U\to \mathbb{C}, K(z,w),z,w\in U with the following properties:

1. K(z,w) is holomorphic in z and anti-holomorphic in w.

   
   K(w,z)=\overline{K(z,w)}
   

2. For each fixed z\in U, K(z,w) is a square integrable d\alpha(w). For all F\in \mathcal{H}L^2(U,\alpha),

   
   F(z)=\int_U K(z,w)F(w) \alpha(w) dw
   

3. If F\in L^2(U,\alpha), let PF denote the orthogonal projection of F onto closed subspace \mathcal{H}L^2(U,\alpha). Then

   
   PF(z)=\int_U K(z,w)F(w) \alpha(w) dw
   

4. For all z,u\in U,

   
   \int_U K(z,w)K(w,u) \alpha(w) dw=K(z,u)
   

5. For all z\in U,

   
   |F(z)|^2\leq K(z,z) \|F\|^2_{L^2(U,\alpha)}
   

Proof

For part 1, By Riesz Theorem, the linear functional evaluation at z\in U on \mathcal{H}L^2(U,\alpha) can be represented uniquely as inner product with some \phi_z\in \mathcal{H}L^2(U,\alpha).

F(z)=\langle F,\phi_z\rangle_{L^2(U,\alpha)}=\int_U F(w)\overline{\phi_z(w)} \alpha(w) dw

And assume part 2 is true, then we have

K(z,w)=\overline{\phi_z(w)}

So part 1 is true.

For part 2, we can use the same argument

\phi_z(w)=\langle \phi_z,\phi_w\rangle_{L^2(U,\alpha)}=\overline{\langle \phi_w,\phi_z\rangle_{L^2(U,\alpha)}}=\overline{\phi_w(z)}

... continue if needed.

Construction of reproducing kernel

Let \{e_j\} be any orthonormal basis of \mathcal{H}L^2(U,\alpha). Then for all z,w\in U,

\sum_{j=1}^{\infty} |e_j(z)\overline{e_j(w)}|<\infty

and

K(z,w)=\sum_{j=1}^{\infty} e_j(z)\overline{e_j(w)}

Bargmann space

The Bargmann spaces are the holomorphic spaces

\mathcal{H}L^2(\mathbb{C}^d,\mu_t)

where

\mu_t(z)=(\pi t)^{-d}\exp(-|z|^2/t)

For this research, we can tentatively set t=1 and d=2 for simplicity so that you can continue to read the next section.

Reproducing kernel for Bargmann space

For all d\geq 1, the reproducing kernel of the space \mathcal{H}L^2(\mathbb{C}^d,\mu_t) is given by

K(z,w)=\exp(z\cdot \overline{w}/t)

where z\cdot \overline{w}=\sum_{k=1}^d z_k\overline{w_k}.

This gives the pointwise bounds

|F(z)|^2\leq \exp(\|z\|^2/t) \|F\|^2_{L^2(\mathbb{C}^d,\mu_t)}

For all F\in \mathcal{H}L^2(\mathbb{C}^d,\mu_t), and z\in \mathbb{C}^d.

Proofs are intentionally skipped, you can refer to the lecture notes for details.

Lie bracket of vector fields

Let X,Y be two vector fields on a smooth manifold M. The Lie bracket of X and Y is an operator [X,Y]:C^\infty(M)\to C^\infty(M) defined by

[X,Y](f)=X(Y(f))-Y(X(f))

This operator is a vector field.

Complex Manifolds

This section extends from our previous discussion of smooth manifolds in Math 401, R2.

For this week [10/21/2025], our goal is to understand the Riemann-Roch theorem and its applications.

References:

• Introduction to Complex Manifolds

Riemann-Roch Theorem (Theorem 9.64)

Suppose M is a connected compact Riemann surface of genus g, and L\to M is a holomorphic line bundle. Then

\dim \mathcal{O}(M;L)=\deg L+1-g+\dim \mathcal{O}(M;K\otimes L^*)